Question

Solve each equation in Exercises 83–108 by the method of your choice. (2x - 5)(x + 1) = 2

Accepted Answer

Rewrite the equation in standard form by expanding the left-hand side and moving all terms to one side of the equation. Start by distributing the terms: $$(2x - 5)(x + 1) = 2. $$Expand $$(2x - 5)(x + 1) to $$get $$2x^2 + 2x - 5x - 5 = 2. $$Combine like terms to simplify: $$2x^2 - 3x - 5 = 2. $$Subtract 2 from both sides to set the equation to zero: $$2x^2 - 3x - 7 = 0. $$Identify the type of equation. This is a quadratic equation in the form $$ax^2 + bx + c = 0$$, where $$a = 2$$, $$b = -3$$, and $$c = -7. $$Choose a method to solve the quadratic equation. You can use factoring (if possible), completing the square, or the quadratic formula. Since factoring may not be straightforward here, the quadratic formula is a good choice: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$Substitute the values of $$a$$, $$b$$, and $$c$$ into the quadratic formula. Replace $$a = 2$$, $$b = -3$$, and $$c = -7$$ into $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$This becomes $$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-7)}}{2(2)}. $$Simplify the expression under the square root (the discriminant) and the rest of the formula. Compute $$(-3)^2 - 4(2)(-7)$$ and simplify the denominator $$2(2). $$Then, simplify the entire expression to find the two possible solutions for $$x.$$

Solve each equation in Exercises 83–108 by the method of your choice. (2x - 5)(x + 1) = 2

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Key Concepts

Factoring

Zero Product Property

Isolating the Variable